Graph this system of equations and solve. $y = \dfrac{1}{3} x - 1$ $y = -\dfrac{4}{3} x + 4$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $\dfrac{1}{3}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $3$ positions to the right. $3$ positions to the right. Graph the blue line so it passes through $(0, -1)$ and $(3, 0)$ The y-intercept for the second equation is $4$ , so the second line must pass through the point $(0, 4)$ The slope for the second equation is $-\dfrac{4}{3}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) You must also move $3$ positions to the right. $3$ positions to the right. $4$ positions down from $(0, 4)$ is $(3, 0)$ Graph the green line so it passes through $(0, 4)$ and $(3, 0)$ The solution is the point where the two lines intersect. The lines intersect at $(3, 0)$.